{"id":11504,"date":"2022-07-12T22:57:01","date_gmt":"2022-07-12T17:27:01","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11504"},"modified":"2022-07-12T22:57:04","modified_gmt":"2022-07-12T17:27:04","slug":"d-and-e-are-points-on-the-sides-ca-and-cb-respectively-of-a-triangle-abc-right-angled-at-c-prove-that-ae2-bc2-ab2-de2","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/d-and-e-are-points-on-the-sides-ca-and-cb-respectively-of-a-triangle-abc-right-angled-at-c-prove-that-ae2-bc2-ab2-de2\/","title":{"rendered":"D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that \\({AE}^2\\) + \\({BC}^2\\) = \\({AB}^2\\) + \\({DE}^2\\)."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>From triangle ACE,<img decoding=\"async\" loading=\"lazy\" class=\" wp-image-11521 alignright\" src=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-13.jpeg?resize=160%2C170&#038;ssl=1\" alt=\"triangle\" width=\"160\" height=\"170\" srcset=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-13.jpeg?resize=282%2C300&amp;ssl=1 282w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-13.jpeg?resize=963%2C1024&amp;ssl=1 963w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-13.jpeg?resize=768%2C817&amp;ssl=1 768w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-13.jpeg?resize=1444%2C1536&amp;ssl=1 1444w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-5-13.jpeg?w=1700&amp;ssl=1 1700w\" sizes=\"(max-width: 160px) 100vw, 160px\" data-recalc-dims=\"1\" \/><\/p>\n<p>\\({AE}^2\\) = \\({EC}^2\\) + \\({AC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0&#8230;&#8230;&#8230;.(1)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (By Pythagoras Theorem)<\/p>\n<p>From triangle DCB,<\/p>\n<p>\\({BD}^2\\) = \\({BC}^2\\) + \\({DC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 &#8230;&#8230;&#8230;(2)<\/p>\n<p>Adding (1) and (2), we get<\/p>\n<p>\\({AE}^2\\) + \\({BD}^2\\) = \\({EC}^2\\) + \\({AC}^2\\) + \\({BC}^2\\) + \\({DC}^2\\)<\/p>\n<p>By Pythagoras Theorem in right triangle ECD and ABC, \\({DE}^2\\) = \\({EC}^2\\) + \\({DC}^2\\)\u00a0 and\u00a0 \\({AB}^2\\) = \\({BC}^2\\) + \\({AC}^2\\)<\/p>\n<p>Hence,\u00a0 <strong>\\({AE}^2\\) + \\({BC}^2\\) = \\({AB}^2\\) + \\({DE}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : From triangle ACE, \\({AE}^2\\) = \\({EC}^2\\) + \\({AC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0&#8230;&#8230;&#8230;.(1)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (By Pythagoras Theorem) From triangle DCB, \\({BD}^2\\) = \\({BC}^2\\) + \\({DC}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 &#8230;&#8230;&#8230;(2) Adding (1) and (2), we get \\({AE}^2\\) + \\({BD}^2\\) = \\({EC}^2\\) + \\({AC}^2\\) + \\({BC}^2\\) + \\({DC}^2\\) By Pythagoras Theorem in right triangle &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/d-and-e-are-points-on-the-sides-ca-and-cb-respectively-of-a-triangle-abc-right-angled-at-c-prove-that-ae2-bc2-ab2-de2\/\"> <span class=\"screen-reader-text\">D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that \\({AE}^2\\) + \\({BC}^2\\) = \\({AB}^2\\) + \\({DE}^2\\).<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,912],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that \\({AE}^2\\) + \\({BC}^2\\) = \\({AB}^2\\) + \\({DE}^2\\). - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/d-and-e-are-points-on-the-sides-ca-and-cb-respectively-of-a-triangle-abc-right-angled-at-c-prove-that-ae2-bc2-ab2-de2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that \\({AE}^2\\) + \\({BC}^2\\) = \\({AB}^2\\) + \\({DE}^2\\). - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : From triangle ACE, ({AE}^2) = ({EC}^2) + ({AC}^2)\u00a0 \u00a0 \u00a0 \u00a0&#8230;&#8230;&#8230;.(1)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (By Pythagoras Theorem) From triangle DCB, ({BD}^2) = ({BC}^2) + ({DC}^2)\u00a0 \u00a0 \u00a0 \u00a0 &#8230;&#8230;&#8230;(2) Adding (1) and (2), we get ({AE}^2) + ({BD}^2) = ({EC}^2) + ({AC}^2) + ({BC}^2) + ({DC}^2) By Pythagoras Theorem in right triangle &hellip; D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that ({AE}^2) + ({BC}^2) = ({AB}^2) + ({DE}^2). 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Prove that \\({AE}^2\\) + \\({BC}^2\\) = \\({AB}^2\\) + \\({DE}^2\\). - Mathemerize","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/mathemerize.com\/d-and-e-are-points-on-the-sides-ca-and-cb-respectively-of-a-triangle-abc-right-angled-at-c-prove-that-ae2-bc2-ab2-de2\/","og_locale":"en_US","og_type":"article","og_title":"D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. 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